Assert a model

\(H_0\): I have \(N_{pairs}\) pairs of socks and \(N_{singles}\) singletons. The first 11 socks that I pull out of the machine are a random sample from this population.

Decide on a test statistic

The number of singletons in the sample: 11.

Construct the sampling distribution

Probability theory or simulation.

See where your observed stat lies in that distribution

Find the p-value if you like.

We'll use simulation.

Create the population of socks:

sock_pairs <- c("A", "B", "C", "D", "E", "F", "G", "H", "I", "J", "K")
sock_singles <- c("l", "m", "n", "o", "p")
socks <- c(rep(sock_pairs, each = 2), sock_singles)
socks

##  [1] "A" "A" "B" "B" "C" "C" "D" "D" "E" "E" "F" "F" "G" "G" "H" "H" "I"
## [18] "I" "J" "J" "K" "K" "l" "m" "n" "o" "p"

picked_socks <- sample(socks, size = 11, replace = FALSE)
picked_socks

##  [1] "J" "D" "n" "F" "p" "B" "m" "l" "H" "I" "C"

sock_counts <- table(picked_socks)
sock_counts

## picked_socks
## B C D F H I J l m n p 
## 1 1 1 1 1 1 1 1 1 1 1

n_singles <- sum(sock_counts == 1)
n_singles

## [1] 11

pick_socks(N_pairs = 9, N_singles = 5, N_pick = 11)

## [1] 9

pick_socks(9, 5, 11)

## [1] 7

pick_socks(9, 5, 11)

## [1] 7

Repeat many, many times…

Quantifying how far into the tails our observed count was.

table(sim_singles)

## sim_singles
##   1   3   5   7   9  11 
##   2  48 248 411 250  41

table(sim_singles)[6]/1000

##    11 
## 0.041

Our two-tailed p-value is 0.082.

The result of a hypothesis test is a probability of the form:

\[ P(\textrm{ data or more extreme } | \ H_0 \textrm{ true }) \]

while most people think they're getting

\[ P(\ H_0 \textrm{ true } | \textrm{ data }) \]

How can we go from the former to the latter?

\[P(A \ | \ B) = \frac{P(A \textrm{ and } B)}{P(B)} \]

\[P(A \ | \ B) = \frac{P(B \ | \ A) \ P(A)}{P(B)} \]

\[P(model \ | \ data) = \frac{P(data \ | \ model) \ P(model)}{P(data)} \]

What does it mean to think about \(P(model)\)?

head(sock_sim)

##   unique pairs n_socks prop_pairs
## 1      3     4      16      0.970
## 2      7     2      33      0.914
## 3      9     1      51      0.929
## 4      1     4       9      0.955
## 5      9     1      45      0.851
## 6      9     1      21      0.726

sock_sim %>%
  filter(unique == 11, pairs == 0) %>%
  head()

##   unique pairs n_socks prop_pairs
## 1     11     0      49      0.692
## 2     11     0      37      0.873
## 3     11     0      49      0.815
## 4     11     0      62      0.961
## 5     11     0      53      0.974
## 6     11     0      59      0.847

• Distribution of a parameter after conditioning on the data
• Synthesis of prior knowledge and observations (data)

Question

What is your best guess for the number of socks that Karl has?

Bayesian methods . . .

• Require the subjective specification of your prior knowledge
• Provide a posterior distribution on the parameters
• Have strong intuition
• Are computationally expensive